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24x^2-160=0
a = 24; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·24·(-160)
Δ = 15360
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{15}}{2*24}=\frac{0-32\sqrt{15}}{48} =-\frac{32\sqrt{15}}{48} =-\frac{2\sqrt{15}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{15}}{2*24}=\frac{0+32\sqrt{15}}{48} =\frac{32\sqrt{15}}{48} =\frac{2\sqrt{15}}{3} $
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